A ball is thrown vertically upwards. It was observed, at a height $$h$$ twice with a time interval $$\Delta t.$$ The initial velocity of the ball is-

Solution :
$$\eqalign{ & h = ut - \frac{1}{2}g{t^2} \cr & \Rightarrow g{t^2} - 2ut + 2h = 0 \cr} $$
solving for $$t$$ we get
$$\eqalign{ & {t_1} + {t_2} = \frac{{2u}}{g} \cr & {t_1} \times {t_2} = \frac{{2h}}{g} \cr & {\text{so,}}\,\Delta t = \left| {{t_1} - {t_2}} \right| = {\left( {{t_1} + {t_2}} \right)^2} - 4{t_1}{t_2} \cr} $$
putting value we get
$$u = \frac{1}{2}\sqrt {8gh + {g^2}\Delta {t^2}} $$